Let the locus of the centre $$(\alpha, \beta)$$, $$\beta > 0$$, of the circle which touches the circle $$x^2 + (y - 1)^2 = 1$$ externally and also touches the $$x$$-axis be $$L$$. Then the area bounded by $$L$$ and the line $$y = 4$$ is

We need to find the area bounded by the locus $$ L $$ of the centre $$ (\alpha, \beta) $$ (with $$ \beta > 0 $$) of a circle that touches $$ x^2 + (y-1)^2 = 1 $$ externally and also touches the $$ x $$-axis, and the line $$ y = 4 $$.

Let the moving circle have centre $$ (\alpha, \beta) $$ and radius $$ r $$. Since it touches the $$ x $$-axis:

$$r = \beta \quad (\text{since } \beta > 0)$$

Since it touches the circle $$ x^2 + (y-1)^2 = 1 $$ (centre $$ (0, 1) $$, radius 1) externally:

$$\sqrt{\alpha^2 + (\beta - 1)^2} = r + 1 = \beta + 1$$

Squaring both sides:

$$\alpha^2 + (\beta - 1)^2 = (\beta + 1)^2$$

$$\alpha^2 + \beta^2 - 2\beta + 1 = \beta^2 + 2\beta + 1$$

$$\alpha^2 = 4\beta$$

Replacing $$ \alpha $$ with $$ x $$ and $$ \beta $$ with $$ y $$, the locus $$ L $$ is:

$$x^2 = 4y$$

This is an upward-opening parabola with vertex at the origin.

When $$ y = 4 $$: $$ x^2 = 16 $$, so $$ x = \pm 4 $$.

The area between the parabola and the line $$ y = 4 $$ is:

$$A = \int_{-4}^{4} \left(4 - \frac{x^2}{4}\right) dx = 2\int_{0}^{4} \left(4 - \frac{x^2}{4}\right) dx$$

$$= 2\left[4x - \frac{x^3}{12}\right]_0^4 = 2\left(16 - \frac{64}{12}\right) = 2\left(16 - \frac{16}{3}\right)$$

$$= 2 \cdot \frac{48 - 16}{3} = 2 \cdot \frac{32}{3} = \frac{64}{3}$$

The area bounded by $$ L $$ and the line $$ y = 4 $$ is $$ \dfrac{64}{3} $$, which corresponds to Option C.